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%I #20 May 10 2014 14:06:18
%S 1,2,2,2,4,4,6,4,6,12,8,12,20,12,12,20,40,24,24,40,70,40,36,40,70,140,
%T 80,72,80,140,252,140,120,120,140,252,504,280,240,240,280,504,924,504,
%U 420,400,420,504,924,1848,1008,840,800,840,1008,1848,3432,1848,1512,1400,1400,1512,1848
%N Irregular triangular array read by rows: T(n,k) is the number of n-step walks (steps +1,-1) on the x-axis beginning at the origin that are on the origin for the last time on step 2k, n>=0, 0<=k<=floor(n/2).
%C Column k=0 is A063886.
%C Row sums give A000079.
%H Alois P. Heinz, <a href="/A237520/b237520.txt">Rows n = 0..200, flattened</a>
%F G.f. for column k: binomial(2k,k) x^k*A(x) where A(x) is the o.g.f. for A063886.
%e 1;
%e 2;
%e 2, 2;
%e 4, 4;
%e 6, 4, 6;
%e 12, 8, 12;
%e 20, 12, 12, 20;
%e 40, 24, 24, 40;
%e 70, 40, 36, 40, 70;
%e 140, 80, 72, 80, 140;
%e 252, 140, 120, 120, 140, 252;
%e T(4,1) = 4 because we have: (-1,+1,-1,-1), (-1,+1,+1,+1), (+1,-1,-1,-1), (+1,-1,+1,+1). These walks have 4 steps and are on the origin for the last time on step 2*1=2.
%p T:= (n, k)-> 2^irem(n, 2)*binomial(2*k, k)*
%p binomial(2*iquo(n, 2)-2*k, iquo(n, 2)-k):
%p seq(seq(T(n, k), k=0..iquo(n, 2)), n=0..14); # _Alois P. Heinz_, May 10 2014
%t nn=20;d=(1-(1-4x^2)^(1/2))/(2x^2);Map[Select[#,#>0&]&,Transpose[Table[ CoefficientList[Series[Binomial[2n,n]x^(2n)(1-2x^2d)/(1-2x),{x,0,nn}],x],{n,0,nn/2}]]]//Grid
%t (* or *)
%t f[list_]:=If[Max[Flatten[Position[list,0]]]== -Infinity,0,Max[Flatten[ Position[list,0]]]]; Table[Distribution[Map[f,Map[Accumulate, Strings[{-1,1},n]]]],{n,0,10}]//Grid
%Y Cf. A067804.
%K nonn,tabf,walk
%O 0,2
%A _Geoffrey Critzer_, Feb 08 2014
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